 Suppose that \( X \) is a connected and compact Riemann surface and let \( \displaystyle f : X \longrightarrow \mathbb{C} \) be a holomorphic function. Show that \( \displaystyle f \) is constant.
 Let \( \displaystyle f : \mathbb{C} \longrightarrow \mathbb{C} \) be a holomorphic bounded function. Show that there is a unique (holomorphic) extension \( \displaystyle \hat{f} : \hat{ \mathbb{C} } \longrightarrow \hat{ \mathbb{C} } \) and then conclude Liouville's theorem (regarding \( \mathbb{C} \)).
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On Riemann Surfaces

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On Riemann Surfaces

 Community Team
 Articles: 0
 Posts: 314
 Joined: Tue Nov 10, 2015 8:25 pm
Re: On Riemann Surfaces
(1) Suppose that \( f \) is not constant. As a nonconstant holomorphic map between Riemann Surfaces, \( f \) is open, so \( f(X) \) is an open subset of \( \mathbb{C} \). Since \( X \) is compact and \( f \) is continuous, \( f(X) \) is also compact and thus a closed subset of \( \mathbb{C} \). But \( \mathbb{C} \) is connected, so \( f(X) = \mathbb{C} \) as \( f \) is not constant. This means that \( \mathbb{C} \) is compact! Contradiction.

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Re: On Riemann Surfaces
(2) Consider the function \( g \) given by \( g(z) = f(\frac{1}{z}) \) on \( \mathbb{C} \smallsetminus \{0\} \). Clearly, \( g \) is holomorphic on \( \mathbb{C} \smallsetminus \{0\} \) and bounded on a neighborhood of \( 0 \), due to the boundedness assumption on \( f \). By Riemann's Removable Singularities Theorem, \( 0 \) is a removable singularity of \( g \), which means that \( g \) can be defined at \( 0 \) in such a way that it becomes holomorphic on all of \( \mathbb{C} \). It follows that \[ \hat{f} \ \colon \hat{\mathbb{C}} \longrightarrow \mathbb{C} \ , \ \hat{f}(z) = \begin{cases} f(z) \ , \ z \in \mathbb{C} \\ g(0) \ , \ z = \infty \end{cases} \]is a holomorphic extension of \( f \) to \( \hat{\mathbb{C}} \). By (1), \( \hat{f} \) is constant and consequently \( f \) itself is constant.